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3 years ago

2x+5y=42 find the slope-intecept?

Mathematics

1 answer:

avanturin [10]3 years ago

4 0

Answer: intercept is 2/5 slope is (0,9)

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The owner of a small store buys coats for $55.00 each. Answer parts a and b. a. She sells the coats for $99.00 each. What perc

scoundrel [369]

Answer:

180% ; 5

Step-by-step explanation:

Purchase price per coat = 55

Sales price per coat = 99

% of purchase price is sales price :

X% * 55 = 99

(x / 100) * 55 = 99

0.01x * 55 = 99

0.55x = 99

x = 99 / 0.55

x = 180

180% of purchase price

B.)

180% * 45 = 81 per jacket

Number of jackets to sell to make atleast $390 in sales

$390 / $81 = 4.8

Hence, 5 jackets

3 0

2 years ago

What is the square of 16

Morgarella [4.7K]

Answer:

Square root of 16 = 4

Square of 16 = 256

Hope this helps :)

Have a great day !

5INGH

Step-by-step explanation:

6 0

3 years ago

Solve it in elimination method 4x+3y=10

zhenek [66]

There is another equation to solve it simultaneously

6 0

3 years ago

the function of y= x^2 - 4x + 8 approximates the height y, of a bird and its horizontal distance, x as it flies from one fence p

Vika [28.1K]

Rewrite the given equation as
$y=x^2-4x+8=x^2-4x+4+4=(x-2)^2+4$ .
Since $(x-2)^2\ge 0$ , you can conclude that the imimum value of y will be gained for the minimum value of $(x-2)^2$ . The minimum value of $(x-2)^2$ is 0 for x=2.
So, y(2)=0+4=4.
Answer: minimum value of y is 4, when x=2.

7 0

2 years ago

Read 2 more answers

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago